A slab of copper is thrust into a parallel-plate capacitor (c26p74) A slab of co

Question

A slab of copper is thrust into a parallel-plate capacitor

(c26p74) A slab of copper of thickness b = 1.379 mm is thrust into a parallel-plate capacitor of C = 7.00×10-11 F of gap d = 8.0 mm, as shown in the figure; it is centered exactly halfway between the plates. What is the capacitance after the slab is introduced?

If a charge q = 9.00×10-6 C is maintained on the plates, what is the ratio of the stored energy before to that after the slab is inserted?

How much work is done on the slab as it is inserted?

Is the slab pulled in or must it be pushed in?

Incorrect pushed
Correct: pulled

Explanation / Answer

Here ,

we can assume that the capacitors are connected in parallel

for the capacitance

Ceq = C * (8 * 2)/(8-1.379)

Ceq = 7 *10^-11 * (8 * 2)/(8-1.379)

Ceq = 1.691 *10^-10 F

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for the ratio of the stroed energy

as ENergy = Q^2 * 0.5/C

ratio of stored energy before to after = Ceq/C

ratio of stored energy before to after = 1.691 *10^-10/(7 *10^-11)

ratio of stored energy before to after = 2.415

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work done in inserting = -0.5 * (9 *10^-6)^2 * (-1/((7 *10^-11) + 1/(1.691 *10^-10))

work done in inserting = -6.84 *10^-21 J

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it must be pulled .

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